Simple Solve - Algebra

Question

kriyen · Answer

$$y= \frac{ 1 }{ \left| x-3\right| } + 2  $$

Find x and y intercepts please.

zzr0ck3r · Answer

what do you get when x=0?

kriyen · Answer

$$\frac{ 7 }{ 3 }$$

zzr0ck3r · Answer

then your y intercept Is (0,7/3)

what do you get when y = 0?

zzr0ck3r · Answer

I would ask my self if the right side could ever be 0

kriyen · Answer

Not sure about that because I'm not sure what working to show with the absolute part?
How do I write out in order to say that it's not possible? 
Or do I just write "not possible" from the beginning ?

zzr0ck3r · Answer

$\frac{1}{|x-3|}+2=0\implies \frac{1}{|x-3|}=-2\implies |x-3|=-\frac{1}{2}$ makes no sense.

kriyen · Answer

Can you elaborate on why it "makes no sense" please =)

zzr0ck3r · Answer

because absolute value is always positive or 0 by definition

zzr0ck3r · Answer

you can think of it as fallows

$|x-3| = \sqrt{(x-3)^2}$

so you have $\sqrt{(x-3)^2}=-\frac{1}{2}\implies (x-3)^2=\sqrt{-\frac{1}{2}}$ $$\huge\text{YIKES}!$$

zzr0ck3r · Answer

the reason I say you may, is because it is not the case that we define $|x| = \sqrt{x^2}$ but it is equivalent on $\mathbb{R}$. In other words, its the same thing.